/* 
 * Median of Two Sorted Arrays
 */

#include "../../func.h"

double findMedianSortedArrays(vector<int> &A, vector<int> &B)
{
    const int m = A.size();
    const int n = B.size();
    int total = m + n;
    if (total & 0x1)
        return find_kth(A.begin(), m , B.begin(), n, total/2+1);
    else
        return (find_kth(A.begin(), m, B.begin(), n, total/2) +
        find_kth(A.begin(), m, B.begin(), n, total/2+1))/2.0;
}

int find_kth(vector<int>::iterator A, int m, vector<int>::iterator B, int n, int k)
{
    if (m > n)
        return find_kth(B, n, A, m, k);
    if (m == 0)
        return *(B + k - 1);
    if (k == 1)
        return min(*A, *B);

    int ia = min(k / 2, m), ib = k - ia;
    if (*(A + ia - 1) < *(B + ib - 1))
        return find_kth(A + ia, m - ia, B, n, k - ia);
    else if (*(A + ia - 1) > *(B + ib - 1))
        return find_kth(A, m, B + ib, n - ib, k - ib);
    else
        return A[ia - 1];
}




/*

void myTest(vector<int> &A, vector<int> &B, int k)
{
    int lA = A.size();
    int lB = B.size();
    int to = lA + lB;

    int sA = 0, sB = 0;
    for (int i = 0; i < to; ++i) {
        if (A[sA] < B[sB] && sA < lA)
            ++sA;
        else if (sB < lB)
            ++sB;
    }
}

double findKth(int a[], int m, int b[], int n, int k)
{
    //always assume that m is equal or smaller than n
    if (m > n)
        return findKth(b, n, a, m, k);
    if (m == 0)
        return b[k - 1];
    if (k == 1)
        return min(a[0], b[0]);
    //divide k into two parts
    int pa = min(k / 2, m), pb = k - pa;
    if (a[pa - 1] < b[pb - 1])
        return findKth(a + pa, m - pa, b, n, k - pa);
    else if (a[pa - 1] > b[pb - 1])
        return findKth(a, m, b + pb, n - pb, k - pb);
    else
        return a[pa - 1];
}
 
class Solution
{
public:
    double findMedianSortedArrays(int A[], int m, int B[], int n)
    {
        int total = m + n;
        if (total & 0x1)
            return findKth(A, m, B, n, total / 2 + 1);
        else
            return (findKth(A, m, B, n, total / 2)
                    + findKth(A, m, B, n, total / 2 + 1)) / 2;
    }
};
*/